# use either product, quotient or chain rule to calcuate d/dx squareroot of 2x+ (3x +4x^2)^3

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Given the function:

f(x)= sqrt(2x) + (3x+4x^2)^3

We need to find d/dx or f'(x).

We will use the chair rule to find the derivative.

f'(x) = [sqrt(2x)]' + [(3x+4x^2)^3]'

We know that sqrt2x= (2x)^1/2 = sqrt2 * (x^1/2)

==> (2x^1/2)' = sqrt2 * (1/2)*x^-1/2 = sqrt2/ 2sqrtx

Now we will differentiate between brackets.

==> Let u= 3x+4x^2

==> du = 3+ 8x d

==> f'(x) = sqrt2/2sqrtx + (u^3)'

= sqrt2/2sqrtx+ 3u^2 du

Now we will substitute:

==> f'(x) = sqrt2/2sqrtx + 3(3x+4x^2)^2 * (3+8x)

= 1/sqrt2x + (9+24x)(3x+4x^2)^2

**==> d/dx = 1/sqrt2x + (9+24x)(3x+4x^2)^2**

We have the function: y = sqrt 2x + (3x + 4x^2)^3

The derivative dy/dx

[sqrt 2x + (3x + 4x^2)^3]'

=> [sqrt 2x]' + [(3x + 4x^2)^3]'

=> [(2x)^(1/2)]' + 3*(3+ 8x)*(3x + 4x^2)^2

=> (1/2)*2(2x)^(-1/2) + 3*(3+ 8x)*(3x + 4x^2)^2

=> 2x^(-1/2) + (9 + 24x)(3x + 4x^2)^2

=> (1/sqrt 2x) + (9 + 24x)(3x + 4x^2)^2

**The required derivative is (1/sqrt 2x) + (9 + 24x)(3x + 4x^2)^2**